Set of algebraic function problems requiring substitution and equation solving.

Let's solve each problem step by step.

---

Problem 6: Given that \( f(x) = 2x^2 - 46 \), solve \( f(x) = 4 \).

#### Step 1: Set up the equation
We are given:
\[ f(x) = 2x^2 - 46 \]
and we need to solve:
\[ f(x) = 4 \]

Substitute \( f(x) = 4 \) into the function:
\[ 2x^2 - 46 = 4 \]

#### Step 2: Solve for \( x \)
1. Add 46 to both sides:
\[ 2x^2 - 46 + 46 = 4 + 46 \]
\[ 2x^2 = 50 \]

2. Divide both sides by 2:
\[ x^2 = 25 \]

3. Take the square root of both sides:
\[ x = \pm \sqrt{25} \]
\[ x = \pm 5 \]

#### Final Answer:
\[ \boxed{x = 5 \text{ or } x = -5} \]

---

Problem 7: A function \( f \) is defined as \( f(x) = x^2 - 9 \).

#### Part (a): Find an expression for \( f(x+5) \)

1. Substitute \( x+5 \) into the function \( f(x) = x^2 - 9 \):
\[ f(x+5) = (x+5)^2 - 9 \]

2. Expand \( (x+5)^2 \):
\[ (x+5)^2 = x^2 + 10x + 25 \]

3. Substitute back into the function:
\[ f(x+5) = x^2 + 10x + 25 - 9 \]
\[ f(x+5) = x^2 + 10x + 16 \]

#### Final Answer for Part (a):
\[ \boxed{x^2 + 10x + 16} \]

#### Part (b): Solve \( f(x+5) = 0 \)

1. From Part (a), we have:
\[ f(x+5) = x^2 + 10x + 16 \]

2. Set \( f(x+5) = 0 \):
\[ x^2 + 10x + 16 = 0 \]

3. Factor the quadratic equation:
\[ x^2 + 10x + 16 = (x + 2)(x + 8) \]

4. Solve for \( x \):
\[ (x + 2)(x + 8) = 0 \]
\[ x + 2 = 0 \quad \text{or} \quad x + 8 = 0 \]
\[ x = -2 \quad \text{or} \quad x = -8 \]

#### Final Answer for Part (b):
\[ \boxed{x = -2 \text{ or } x = -8} \]

---

Problem 8: Given that \( f(x) = x^2 - 1 \), solve \( f(x-3) = 0 \).

#### Step 1: Substitute \( x-3 \) into the function
We are given:
\[ f(x) = x^2 - 1 \]
and we need to solve:
\[ f(x-3) = 0 \]

Substitute \( x-3 \) into the function:
\[ f(x-3) = (x-3)^2 - 1 \]

#### Step 2: Solve for \( x \)
1. Set \( f(x-3) = 0 \):
\[ (x-3)^2 - 1 = 0 \]

2. Add 1 to both sides:
\[ (x-3)^2 = 1 \]

3. Take the square root of both sides:
\[ x-3 = \pm 1 \]

4. Solve for \( x \):
\[ x-3 = 1 \quad \text{or} \quad x-3 = -1 \]
\[ x = 4 \quad \text{or} \quad x = 2 \]

#### Final Answer:
\[ \boxed{x = 4 \text{ or } x = 2} \]

---

Problem 9: Given that \( f(x) = 5x + 1 \) and \( g(x) = 3x - 7 \), solve \( f(x) = g(x) \).

#### Step 1: Set the functions equal to each other
We are given:
\[ f(x) = 5x + 1 \]
\[ g(x) = 3x - 7 \]

Set \( f(x) = g(x) \):
\[ 5x + 1 = 3x - 7 \]

#### Step 2: Solve for \( x \)
1. Subtract \( 3x \) from both sides:
\[ 5x + 1 - 3x = 3x - 7 - 3x \]
\[ 2x + 1 = -7 \]

2. Subtract 1 from both sides:
\[ 2x + 1 - 1 = -7 - 1 \]
\[ 2x = -8 \]

3. Divide both sides by 2:
\[ x = \frac{-8}{2} \]
\[ x = -4 \]

#### Final Answer:
\[ \boxed{x = -4} \]

---

Final Answers Summary:

1. Problem 6: \( \boxed{x = 5 \text{ or } x = -5} \)
2. Problem 7(a): \( \boxed{x^2 + 10x + 16} \)
3. Problem 7(b): \( \boxed{x = -2 \text{ or } x = -8} \)
4. Problem 8: \( \boxed{x = 4 \text{ or } x = 2} \)
5. Problem 9: \( \boxed{x = -4} \)